Alternating Current — Page 10 | JEE Physics

Q91

A circuit connected to an ac source of emf e = e0sin(100t) with t in seconds, gives a phase difference of

\pi

/4 between the emf e and current i. Which of the following circuits will exhibit this ?

A RC circuit with R = 1 k

\Omega

and C = 1μF

B RL circuit with R = 1k

\Omega

and L = 1mH

C RC circuit with R = 1k

\Omega

and C = 10 μF

D RL circuit with R = 1 k

\Omega

and L = 10 mH

Correct Answer

Option C

Solution

Given phase difference =

{\pi \over 4}

and $\omega$ = 100 rad/s $\Rightarrow$ Reactance (X) = Resistance (R) Now by checking option. Option (A) R = 1000

\Omega

and Xc =

{1 \over {{{10}^{ - 6}} \times 100}} = {10^4}\Omega

Option (B) R = 103

\Omega

and XL =

{10^{ - 3}} \times 100 = 10^{-1} \Omega

Option (C) R = 103

\Omega

and Xc =

{1 \over {{10 \times {10}^{ - 6}} \times 100}} = {10^3}\Omega

Option (D) R = 103

\Omega

and XL =

10 \times {10^{ - 3}} \times 100 = 1\Omega

Q92

An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5

\Omega

resistor. The ratio of the currents at time t =

\infty

and at t = 40 s is close to : (Take e2 = 7.389)

A 1.06

B 0.84

C 1.15

D 1.46

Correct Answer

Option A

Solution

i = i0(1 -

{e^{ - {{Rt} \over L}}}

) i $\infty$ = i0(1 -

{e^{ - \infty }}

) = i0 $\therefore$

{{{i_\infty }} \over {{i_{40s}}}}

=

{{{i_0}} \over {{i_0}\left( {1 - {e^{ - {{5 \times 40} \over {10 \times {{10}^{ - 3}}}}}}} \right)}}

=

{1 \over {\left( {1 - {e^{ - 2000}}} \right)}}

$\approx$ 1 Then most appropriate option is 1.06

Q93

An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to :

A 0.044 H

B 0.065 H

C 80 H

D 0.08 H

Correct Answer

Option B

Solution

From the circuit, we have

R = {{80} \over {10}} = 8\,\Omega

10 = {{220} \over {\sqrt {{R^2} + X_L^2} }} \Rightarrow \sqrt {64 + X_L^2} = 22

\Rightarrow X_L^2 = 484 - 64 = 420

\Rightarrow {X_L} = \sqrt {420} = 20.5\,\Omega

\Rightarrow \omega L = 20.5

L = {{20.5} \over {2\pi \times 50}} = {{20.5} \over {314}} = 0.065\,H

Q94

An AC voltage

V=20 \sin 200 \pi t

is applied to a series LCR circuit which drives a current

I=10 \sin \left(200 \pi t+\dfrac{\pi}{3}\right)

. The average power dissipated is:

A 21.6 W

B 200 W

C 173.2 W

D 50 W

Correct Answer

Option D

Solution

\begin{aligned} & =\mathrm{IV} \cos \phi \\ & =\frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times \cos 60^{\circ} \\ & =50 \mathrm{~W} \end{aligned}

Q95

A series LCR circuit driven by 300 V at a frequency of 50 Hz contains a resistance R = 3 k

\Omega

, an inductor of inductive reactance XL = 250

\pi

\Omega

and an unknown capacitor. The value of capacitance to maximize the average power should be : (Take

\pi

2 = 10)

A 4

\mu

F

B 25

\mu

F

C 400

\mu

F

D 40

\mu

F

Correct Answer

Option A

Solution

From maximum average power XL = XC 250 $\pi$ =

{1 \over {2\pi (50)C}}

$\Rightarrow$ C = 4 $\times$ 10 $-$ 6

Q96

A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R=5

\Omega

, L=25 mH and C=1000

\mu

F. The total impedance, and phase difference between the voltage across the source and the current will respectively be :

A 10

\Omega

and tan

-

1

\left( {{5 \over 3}} \right)

B

7\,\Omega

and 45o

C

10\,\Omega

and tan

-

1

\left( {{8 \over 3}} \right)

D

7\,\Omega

and tan

-

1

\left( {{5 \over 3}} \right)

Correct Answer

Option B

Solution

It is given that e0 = 283 V; $\omega$ = 320. The inductor reactance is XL = 320 $\times$ 25 $\times$ 10 $-$ 3 = 8

\Omega

The capacitor reactance is

{X_C} = {1 \over {\omega C}} = {1 \over {320 \times 1000 \times {{10}^{ - 6}}}} = {{1000} \over {320}} = 3.1\,\Omega

It is given that R = 5

\Omega

. Therefore, the total impedance is

Z = \sqrt {{R^2} + {{({X_L} - {X_C})}^2}} = \sqrt {50} = 7\,\Omega

and the phase difference between the voltage across the source and the current is

\tan \phi = {{{X_L} - {X_C}} \over R} = {{8 - 3.1} \over 5} \approx 1 \Rightarrow \theta = 45^\circ

Q97

A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then the time at which the energy stored in the inductor reaches

\left( {{1 \over n}} \right)

times of its maximum value, is :

A

{L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n + 1}}} \right)

B

{L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)

C

{L \over R}\ln \left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)

D

{L \over R}\ln \left( {{{\sqrt n - 1} \over {\sqrt n }}} \right)

Correct Answer

Option B

Solution

P.E. in inductor,

U = {1 \over 2}L{I^2}

U \propto {I^2}

{U \over {{U_0}}} = {\left( {{I \over {{I_0}}}} \right)^2}

{1 \over n} = {\left( {{I \over {{I_0}}}} \right)^2}

I = {{{I_0}} \over {\sqrt n }}

We know,

I = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)

{{{I_0}} \over {\sqrt n }} = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)

$\Rightarrow$

{e^{ - {{Rt} \over L}}}

= 1 -

{1 \over {\sqrt n }}

taking ln & solving we get,

t = {L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)

Q98

The phase difference between the alternating current and

emf

is

{\pi \over 2}.

Which of the following cannot be the constituent of the circuit?

A

R,L

B

C

alone

C

L

alone

D

L, C

Correct Answer

Option A

Solution

The phase difference between the alternating current and emf in an AC circuit depends on the components in the circuit: In a purely resistive ( $R$ ) circuit, the current and emf are in phase, meaning the phase difference is $0$ .

In a purely inductive ( $L$ ) circuit, the current lags behind the emf by $\dfrac{\pi}{2}$ , meaning the phase difference is $\dfrac{\pi}{2}$ .

In a purely capacitive ( $C$ ) circuit, the current leads the emf by $\dfrac{\pi}{2}$ , again meaning the phase difference is $\dfrac{\pi}{2}$ .

In an $L$ - $R$ or $L$ - $C$ circuit, the phase difference depends on the relative values of $L$ , $R$ , and $C$ and can be anywhere between $0$ and $\dfrac{\pi}{2}$ .

Therefore, if the phase difference between the alternating current and emf is $\dfrac{\pi}{2}$ , then the circuit cannot contain only a resistor ( $R$ ) since that would give a phase difference of $0$ .

So, the answer is Option A: $R,L$ .

The phase difference would not be $\dfrac{\pi}{2}$ if the circuit contains both a resistor and an inductor.

Q99

A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant 'b', the correct equivalence would be:

A L

\leftrightarrow

k, C

\leftrightarrow

b, R

\leftrightarrow

m

B L

\leftrightarrow

m, C

\leftrightarrow

k, R

\leftrightarrow

b

C L

\leftrightarrow

m, C

\leftrightarrow

{1 \over k}

, R

\leftrightarrow

b

D L

\leftrightarrow

{1 \over b}

, C

\leftrightarrow

{1 \over m}

, R

\leftrightarrow

{1 \over k}

Correct Answer

Option C

Solution

For spring mass damped oscillator ma = - kx - bv $\Rightarrow$ ma + kx + bv = 0 $\Rightarrow$

m{{{d^2}x} \over {d{t^2}}}

+ b

{{dx} \over {dt}}

+ kx = 0 ....(1) For LCR circuit L

{{di} \over {dt}}

+ iR +

{q \over C}

= 0 $\Rightarrow$ L

{{{d^2}q} \over {d{t^2}}}

+ R

{{dq} \over {dt}}

+

{q \over C}

= 0 .....(2) Comparing (1) and (2), we get L $\leftrightarrow$ m, C $\leftrightarrow$

{1 \over k}

, R $\leftrightarrow$ b